| ⇦ |
| ⇨ |
A Carnot engine, having an efficiency of η=1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is
Options
(a) 100 J
(b) 99 J
(c) 90 J
(d) 1 J
Correct Answer:
90 J
Explanation:
Efficiency of carnot engine n = 1 – (T₂ / T₁)
That is, 1 / 10 = 1 – (T₂ / T₁)
⇒ (T₂ / T₁) = 1 – (1 / 10) = 9 / 10 ⇒ (T₁ / T₂) = 10 / 9
.·. w = Q₂ . [(T₁ / T₂) – 1) ⇒ 10 = Q₂ [(10 / 9) – 1]
⇒ 10 = Q₂ (1 / 9) ⇒ Q₂ = 90 J
So, 90 J heat is absorbed at lower temperature.
Related Questions: - An electromagnetic wave of frequency v = 3.0 MHz passes from vacuum into
- Two sources P and Q produce notes of frequency 660 Hz each.
- In the diffraction pattern of a single slit
- A particle of mass m is projected with velocity v making an angle of 45° with
- A ship A is moving westwards with a speed of 10 km/h .The time after
Topics: Thermodynamics
(179)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- An electromagnetic wave of frequency v = 3.0 MHz passes from vacuum into
- Two sources P and Q produce notes of frequency 660 Hz each.
- In the diffraction pattern of a single slit
- A particle of mass m is projected with velocity v making an angle of 45° with
- A ship A is moving westwards with a speed of 10 km/h .The time after
Topics: Thermodynamics (179)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply